Let $a,b,c>0: abc=1.$ Prove that: $$2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a). $$
I've tried to use a well-known lemma but the rest is quite complicated for me.
Lemma. For any positive real numbers $x,y,z: xyz=1$ then $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\sqrt{3(x^2+y^2+z^2)}.$$
Proof for lemma.
Obviously, we need prove that $$\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}+2\left(\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\right)\ge3(x^2+y^2+z^2).$$
By AM-GM $$\frac{x^2}{y^2}+\frac{x}{z}+\frac{x}{z}\ge3\sqrt[3]{\frac{x^4}{y^2z^2}}=3x^2,$$
$$\frac{y^2}{z^2}+\frac{y}{x}+\frac{y}{x}\ge3\sqrt[3]{\frac{y^4}{x^2z^2}}=3y^2,$$
and
$$\frac{z^2}{x^2}+\frac{z}{y}+\frac{z}{y}\ge3\sqrt[3]{\frac{z^4}{y^2x^2}}=3z^2.$$
Thus, the lemma is proven. Equality holds iff $x=y=z=1.$
Back on the main problem, by applying the lemma, it's enough to prove $$2(a+b+c)\left(1+\sqrt{3(a^2+b^2+c^2)}\right)\ge 3(a+b)(b+c)(c+a). $$
I stopped here. It seems that we can square but it's quite complicated.
Hope you help me prove my last inequality. Thank you for your help.
Best Answer
Proof.
Firstly, rewrite the inequality as: $$2(a+b+c)(ab+bc+ca)(a^2c+c^2b+b^2a+1)\ge3(a+b)(b+c)(c+a)(ab+bc+ca)$$, $$\iff \frac{1+a^2c+c^2b+b^2a}{(a+b)(b+c)(c+a)}+1+a^2c+c^2b+b^2a\ge\frac{3(ab+bc+ca)}{2},$$ $$\iff 1+a^2c+c^2b+b^2a+\frac{a(a+b)(b+c)+b(b+c)(c+a)+c(c+a)(a+b)-(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}\ge\frac{3(ab+bc+ca)}{2},$$ $$\iff \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}\ge\frac{3(ab+bc+ca)}{2},$$ Or $$\frac{a}{b(a+c)}+\frac{b}{c(b+a)}+\frac{c}{a(c+b)}\ge \frac{3(ab+bc+ca)}{2(a+b+c)}. \tag{*}$$ From here, we can use the lemma you proved.
Back on main problem, using C-S inequality and the lemma for (**): $$\frac{\dfrac{ac}{bc}}{a+c}+\frac{\dfrac{ba}{ca}}{a+b}+\frac{\dfrac{cb}{ab}}{b+c}\ge\frac{\left(\sqrt{\dfrac{ab}{bc}}+\sqrt{\dfrac{bc}{ca}}+\sqrt{\dfrac{ca}{ab}}\right)^2}{2(a+b+c)}\ge\frac{3(ab+bc+ca)}{2(a+b+c)}$$The problem is completely proven. Equality holds iff: $a=b=c=1.$